Problem 1Given, |
> y := 3*x^3-2*x^2+5;
3 2
y := 3 x - 2 x + 5
|
a) Find the length from x=0 to x=1. b) Find the curvature at x=1/2. c) Give a parametrization as a space curve. d) Compute the unit tangent vector at x=1/2. Sol 1a) the length is: |
> lth := Int(sqrt(1+ (diff(y,x))^2),x=0..1);
1
/
| 4 3 2 1/2
lth := | (1 + 81 x - 72 x + 16 x ) dx
|
/
0
> sol_a := evalf(lth);
sol_a := 1.796448727
| b) For the curvature we use the formula when y=f(x). |
> k = diff(diff(f(x),x),x)/ (1+diff(f(x),x)^2)^(3/2);
2
d
----- f(x)
2
dx
k = ---------------------
/ / d \2\3/2
|1 + |---- f(x)| |
\ \ dx / /
> y1 := diff(y,x); y2 := diff(y1,x); k := y2/(1+y1^2)^(3/2);
2
y1 := 9 x - 4 x
y2 := 18 x - 4
18 x - 4
k := ----------------------
2 2 3/2
(1 + (9 x - 4 x) )
| So at x=1/2 we get, |
> sol_b := evalf(subs(x=1/2,k));
sol_b := 4.565376471
| c) The parametrization as a space curve can be obtained by letting t=x. |
> r := t -> t*i+(3*t^3-2*t^2+5)*j: 'r(t)' = r(t);
3 2
r(t) = t i + (3 t - 2 t + 5) j
| d) The derivative of r(t) gives a vector tangent to the curve so, |
> sol_d := subs(t=1/2, [1, 9*t^2-4*t]/sqrt(1+(9*t^2+4*t)^2));
1/2 1/2
sol_d := 1/305 [1, 1/4] 305 16
> sol_d := evalf(evalm(sol_d));
sol_d := [ .2290393338, .05725983345 ]
Problem 2The velocity of a particle moving in space is, |
> k := 'k': v := t -> ln(1+t)*i - (t^2+1)*j - t*cos(t)*k: 'v(t)' = v(t);
2
v(t) = ln(1 + t) i - (t + 1) j - t cos(t) k
if at t=0 the particle is at the origin, find the position of the particle at t=1.
Sol 2Since the velocity vector is the derivative w.r. to time of the position vector R then, |
> R := R0 + Int(v(s),s=0..t);
t
/
| 2
R := R0 + | ln(1 + s) i - (s + 1) j - s cos(s) k ds
|
/
0
| where R0 is, the position at t=0. We have |
> R0 := 0: i:= 'i': j:='j':k:='k': R := subs(t=1,R);
1
/
| 2
R := | ln(1 + s) i - (s + 1) j - s cos(s) k ds
|
/
0
> R := evalf(int(ln(1+s),s=0..1))*i-evalf(int(s^2+1,s=0..1))*j - R := .386294361 i - 1.333333333 j - .381773291 k
Problem 3Consider the function: |
> x := 'x':y:='y':
> f := (x,y) -> (x^2+y^2-2*x-2*y)/(x^2+y^2-2*x+2*y+2): 'f(x,y)' = f(x,y);
2 2
x + y - 2 x - 2 y
f(x, y) = -----------------------
2 2
x + y - 2 x + 2 y + 2
| Find the limit at (1,-1) it it exists. i.e., |
> #
2 2
x + y - 2 x - 2 y
Limit -----------------------
(x,y) ->(1,-1) 2 2
x + y - 2 x + 2 y + 2
Sol 3If the function is continuous at (1,-1) all we need is to plug-in x=1 and y=-1 so let us first check if the function is at least defined at (1,-1) |
> f(1,-1);
Error, (in f) division by zero
| So it is not continuous. The denominator is zero there. The limit may still exist however. Notice that the denominator can be factorized as, |
> '(x-1)^2+(y+1)^2' = expand((x-1)^2+(y+1)^2);
2 2 2 2
(x - 1) + (y + 1) = x + y - 2 x + 2 y + 2
| Hence, since the denominator is the sum of two squares it is never negative. If we let r=(x,y) and r0=(1,-1) we have, |
> #
2 2
x + y - 2 x - 2 y
Limit ----------------------- =
r -> r0 2 2
x + y - 2 x + 2 y + 2
/ 2 2 \ / 1 \
|Limit x + y - 2 x - 2 y| * |Limit -------------------|
\r -> r0 / |r -> r0 2 2|
\ (x - 1) + (y + 1) /
this is nothing but the limit of a product is the product of the limits which is true when all the limits exist. So the limit exists and it is 2*infinity
which is just infinity.
Problem 4Given the function, |
> x:='x':y:='y':f:=(x,y)->x*exp(-y)+3*y*cos(x): 'f(x,y)'= f(x,y);
f(x, y) = x exp(- y) + 3 y cos(x)
Find the partial derivative w.r.t. x at (-1,2) and w.r.t. y at (0,1/2). Sol 4We can use either unapply or subs. Let's use both, |
> fx := unapply(diff(f(x,y),x),x,y)(-1,2);
fx := exp(-2) + 6 sin(1)
> fy := subs({x=0,y=1/2},diff(f(x,y),y));
fy := 3 cos(0)
| or approximately |
> fx := evalf(fx); fy := simplify(fy);
fx := 5.184161192
fy := 3
Problem 5Find a normal vector to the plane tangent to the graph of the function, |
> z := sqrt(3*x+y^2);
2 1/2
z := (3 x + y )
at the point (1,-1,2). Sol 5The tangent plane at (x0,y0,z0) has equation, |
> A*(x-x0)+B*(y-y0)-('z'-z0) = 0;
A (x - x0) + B (y - y0) - z + z0 = 0
| where, A and B are the partials w.r.t. x and y resp. at the given point. i.e. |
> A := subs({x=1,y=-1},diff(z,x)); B := subs({x=1,y=-1},diff(z,y)); C:=-1;
1/2
A := 3/8 4
1/2
B := - 1/4 4
C := -1
| so a normal to the plane will be, |
> N := [simplify(A),simplify(B),-1];
N := [3/4, -1/2, -1]
Problem 6Use differentials to approximate, |
> #
2 2 1/2
(9 (1.95) + 8.1 )
Sol 6We first notice that the exact value is f(1.95,8.1) where, |
> f := (x,y) -> sqrt(9*x^2+y^2): 'f(x,y)' = f(x,y);
2 2 1/2
f(x, y) = (9 x + y )
| So the approxiamation is, |
> 'f(1.95,8.1)' = 'f(2,8)'+ 'fx'*dx + 'fy'*dy + bit_more;
f(1.95, 8.1) = f(2, 8) + fx dx + fy dy + bit_more
| where, |
> fx :=simplify(subs({x=2,y=8},diff(f(x,y),x)));
fx := 9/5
> fy :=simplify(subs({x=2,y=8},diff(f(x,y),y)));
fy := 4/5
> dx := 1.95-2; dy := 8.1-8;
dx := -.05
dy := .1
| The approximation is then, |
> apprx := f(2,8)+fx*dx+fy*dy;
apprx := 9.990000000
| The exact value is, |
> exact := f(1.95,8.1);
exact := 9.991621490
Problem 7Let, |
> f :='f': z := f(x,y);
z := f(x, y)
| if x and y are given by, |
> x := r^2+s^2; y := 2*r*s;
2 2
x := r + s
y := 2 r s
| Find, |
> zs := Diff(z,s): Zrs := Diff(zs,r);
2
d z
Zrs := -------
dr ds
Sol 7This is an application of the chain rule. But with maple we can use BRUTE force to get the answer directly: |
> Zrs := simplify(diff(diff(z,s),r));
2 2 2 2 2
Zrs := 4 s D[1, 1](f)(r + s , 2 r s) r + 4 D[1, 2](f)(r + s , 2 r s) s
2 2 2 2 2
+ 4 D[1, 2](f)(r + s , 2 r s) r + 4 r D[2, 2](f)(r + s , 2 r s) s
2 2
+ 2 D[2](f)(r + s , 2 r s)
| where D[1,2](f)(x0,y0) is maple's notation for second order partial of x w.r.t. x and y evaluated at (x0,y0)... etc... |